Calculating Force on Chamber in Water Flow

[stunner5000pt]

Homework Statement

A chamber has water entering horizontally with a velcoity of 80m/s with an inlet of an area of [itex] 0.1m^[/itex] and leaving through an opening of area [itex]0.15m^2[/itex]. The exiting flow makes an angle of 30 degrees with respect to the entering flow. What force is needed to hold the chamber in place

2. The attempt at a solution
Well the total force on the chamber is
[tex] \vec{F}=\dot{m_{2}}\vec{v_{2}}-\dot{m_{1}}\vec{v_{1}} [/tex]
and [tex] \dot{m_{2}}=\rho A_{2} [/tex]
and [tex] \dot{m_{1}}=\rho A_{1} [/tex]

the component of the force is

[tex]F_{x}=\rho A_{2} v_{2} \cos\theta - \rho A_{1} v_{1}[/tex]
[tex]F_{y}=\rho A_{2} v_{2}\sin\theta[/tex]

Now here's the thing, we don't know v2...
But since the mass is conserved the volume flow rate in is equal to the volume flow rate out
[tex]\int_{S_{inlet}} \vec{v_{1}}\cdot\hat{n}dA=\int_{S_{outlet}} \vec{v_{2}}\cdot\hat{n}dA[/tex]
[tex]v_{1}A_{1}=v_{2}A_{2}[/tex]

From this we can find the X and Y components and hence the force and the direction. Is this all correct?

Thank you for your input and suggestions!

 

[Q_Goest]

Hi stunner,

Now here's the thing, we don't know v2...
But since the mass is conserved the volume flow rate in is equal to the volume flow rate out
[tex]\int_{S_{inlet}} \vec{v_{1}}\cdot\hat{n}dA=\int_{S_{outlet}} \vec{v_{2}}\cdot\hat{n}dA[/tex]
[tex]v_{1}A_{1}=v_{2}A_{2}[/tex]

From this we can find the X and Y components and hence the force and the direction. Is this all correct?

Works for me. To be nit picky, the mass flow in = mass flow out (not volume flow) unless there's mass being stored inside the control volume, so we can equate VA(in) = VA(out) only if density(in) = density(out). Note that if this were a gas for example, and density changed, we'd have to determine velocity from the change in density and mass flow. But yea, you got it right.

 

[stunner5000pt]

thanks a lot

you can mark is solved